91Ó°ÊÓ

When dealing with composite functions in calculus, like the function given in the original exercise, the Chain Rule becomes an essential tool for differentiation. Here, the function is initially expressed as \( y = \cos^2 x \), which can be rewritten in a more chain-friendly form, \( y = (\cos x)^2 \). This allows us to identify an "inner" function and an "outer" function.

In simple terms, the Chain Rule states that to differentiate a composite function \( f(g(x)) \), you need to multiply the derivative of the outer function \(f\) (evaluated at the inner function \(g(x)\)) by the derivative of the inner function \(g\). For our specific example:

• Let \( u = \cos x \) (inner function), so \( y = u^2 \) (outer function).
• The derivative of the outer function \( u^2 \) is \( 2u \) and the derivative of the inner function \( \cos x \) is \(-\sin x\).

Applying the Chain Rule, we have \( y' = 2u \cdot \frac{du}{dx} = 2\cos x \cdot (-\sin x) = -2\cos x\sin x \). This becomes \( y' = -\sin(2x) \) using a trig identity.

Trigonometric functions, like \(\sin x\) and \(\cos x\), play a vital role in calculus, especially in problems involving periodic phenomena. These functions have well-known derivatives crucial for solving calculus problems. For instance, the derivative of \(\sin x\) is \(\cos x\), while the derivative of \(\cos x\) is \(-\sin x\).In the initial part of the solution, we approached \(y = \cos^2 x\), and by identifying "\(u = \cos x\)", we used a basic yet frequently applied trig derivative:

• The derivative of \(\cos x\) is \(-\sin x\).

Trigonometric identities can simplify expressions in differentiation. For example, in our exercise, \(-2\cos x \sin x\) was rewritten using the double angle identity \(-\sin(2x)\). Such identities enhance our ability to handle more complex simplifications and derivations.

The second derivative measures how the rate of change of a function's slope itself changes. In other words, it tells us something about the function's acceleration or concavity. Once we have the first derivative, finding the second derivative requires differentiating again.In our problem, after obtaining the first derivative \(y' = -\sin(2x)\), the second derivative \(y''\) is found by differentiating \(-\sin(2x)\):

• Using the chain rule once more, we find \(\frac{d}{dx}[-\sin(2x)] = -\cos(2x) \cdot 2 \).

Thus, the second derivative is calculated as \(y'' = -2 \cos(2x) \). Second derivatives provide understanding about the function's concavity:

• If \(y'' > 0\), the function is concave up.
• If \(y'' < 0\), the function is concave down.

This knowledge is useful for analyzing the behavior and stability of functions in calculus.